1051 Pop Sequence (25 分)

Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., N and pop randomly. You are supposed to tell if a given sequence of numbers is a possible pop sequence of the stack. For example, if M is 5 and N is 7, we can obtain 1, 2, 3, 4, 5, 6, 7 from the stack, but not 3, 2, 1, 7, 5, 6, 4.

Input Specification:

Each input file contains one test case. For each case, the first line contains 3 numbers (all no more than 1000): M (the maximum capacity of the stack), N (the length of push sequence), and K (the number of pop sequences to be checked). Then K lines follow, each contains a pop sequence of N numbers. All the numbers in a line are separated by a space.

Output Specification:

For each pop sequence, print in one line "YES" if it is indeed a possible pop sequence of the stack, or "NO" if not.

Sample Input:

5 7 5
1 2 3 4 5 6 7
3 2 1 7 5 6 4
7 6 5 4 3 2 1
5 6 4 3 7 2 1
1 7 6 5 4 3 2

 

Sample Output:

YES
NO
NO
YES
NO

    题解：使用栈，每压入一个元素就判断是否与当前输出序列值相等，相等就循环弹出，注意压栈前判满和弹出时判空

#include<bits/stdc++.h>
using namespace std;
const int maxn=1010;
#define inf 0x3fffffff
vector<int> ve[maxn];
int main(){
    int m,n,k,temp;
    int A[maxn];
    scanf("%d %d %d",&m,&n,&k);
    for(int i=0;i<k;i++){
        for(int j=0;j<n;j++){
            scanf("%d",&temp);
            ve[i].push_back(temp);
        }
        stack<int> s;
        int r=0;
        bool flag=true;
        for(int t=1;t<=n;t++){
            if(s.size()<m){
                s.push(t);
            }
            else{
                flag=false;
            }
            while(!s.empty()&&s.top()==ve[i][r]){
                r++;
                s.pop();
            }
        }
        if(!s.empty()||r<n){
            printf("NO\n");
        }
        else{
            printf("YES\n");
        }
    }
    return 0;
}

 

标签：1051,sequence,int,NO,pop,Pop,stack,25,numbers
来源： https://www.cnblogs.com/dreamzj/p/15022279.html